Question

Tìm xy: x^2 +10y^2-2xy +6x+1 =0

 

Answer 1

Sửa đề: \(x^2+10y^2-2xy+6y+1=0\)

\(\Leftrightarrow x^2-2xy+y^2+9y^2+6y+1=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(3y+1\right)^2=0\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{3}\)

Answer 2

\(x^2+10y^2-2xy+6x+1=0\Leftrightarrow\left(x-y\right)^2+\left(3y+1\right)^2=0\)

Vì \(\left(x-y\right)^2\ge0,\left(3y+1\right)^2\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}x-y=0\\3y+1=0\end{matrix}\right.\)\(\Rightarrow x=y=\dfrac{-1}{3}\)